原題連結

Given a string S, we can transform every letter individually to be lowercase or uppercase to create another string.  Return a list of all possible strings we could create.

Examples:
Input: S = "a1b2"
Output: ["a1b2", "a1B2", "A1b2", "A1B2"]

Input: S = "3z4"
Output: ["3z4", "3Z4"]

Input: S = "12345"
Output: ["12345"]
Note:

S will be a string with length at most 12.
S will consist only of letters or digits.

 

分析：每個字母可大寫可小寫，用0，1代表該字母的大小寫的狀態，1為大寫。若字串為a1b2，包含兩個字母，則兩個字母所有的狀態組合為{00，01，10，11}，將列舉狀態看作二進位制數，則化為十進位制為{0，1，2，3}，這樣問題就能很好地被解決了。

#include<iostream>
#include<vector>
#include<string>
#include<bitset>
#include<math.h>
using namespace std;

vector<string> letterCasePermutation(string& s)
{
   vector<string> ans;
   vector<int> letterPos;
   
   int mark = 1,size = s.size();
   if(size == 0 ) return ans;

   for(int i = 0; i < size; i++)
   {
       if(isalpha(s[i]))
       {
           s[i] = tolower(s[i]);
           mark *= 2;
           letterPos.push_back(i);
       }
   }

   for(int i = 0; i < mark; i++)
   {
       int t = i;
       string ts = s;
       for(int j = 0;j < letterPos.size(); j++)
       {
           //cout << t << " " << j << " " << (t >> j) % 2 << endl;
           if((t >> j) % 2 == 1) 
           {
               ts[letterPos[letterPos.size() - 1 - j]] = toupper(s[letterPos[letterPos.size() - 1 - j]]);
           }
           //ts[letterPos[j]] = (t >> j) % 2 == 1 ? toupper(s[letterPos[j]]) : s[letterPos[j]];
       }
       cout << ts << endl;
       ans.push_back(ts);

   }

   return ans;
}
int main()
{
   string s;
   s = "a1b2";
   vector<string> ans = letterCasePermutation(s);
   //for(int i = 0;i < ans.size(); i++)  cout << ans[i] << endl;
}